For which positive integers $n$ does there exist a positive real number $\epsilon$ such that
$\dfrac{2^{\omega(n)}}{n^\epsilon}\geq\dfrac{2^{\omega(k)}}{k^\epsilon}$ for all $k<n$, and
$\dfrac{2^{\omega(n)}}{n^\epsilon}>\dfrac{2^{\omega(k)}}{k^\epsilon}$ for all $k>n$?
A divisor $d$ of $n$ is called unitary if $\gcd(d,n/d)=1$, therefore the unitary divisors are precisely the numbers resulting from expanding the product $(p_1^{a_1}+1)\cdots(p_{\omega(n)}^{a_{\omega(n)}}+1)$, where $p_1^{a_1}\cdots p_{\omega(n)}^{a_{\omega(n)}}$ is the canonical prime factorization of $n$. It follows that $n$ has $2^{\omega(n)}$ unitary divisors, compared to $d(n)=(a_1+1)\cdots(a_{\omega(n)}+1)$ divisors unitary or not.
This is of the same form as the definition of a superior highly composite number:
$n$ is called superior highly composite if
$\dfrac{d(n)}{n^\epsilon}\geq\dfrac{d(k)}{k^\epsilon}$ for all $k<n$, and
$\dfrac{d(n)}{n^\epsilon}>\dfrac{d(k)}{k^\epsilon}$ for all $k>n$
Which values satisfy this (the first) definition? I don't know of a method for finding them.
Edit: Every superior highly composite number is a highly composite number, $n$ such that $d(n)>d(k)$ for all $k<n$. Likely the numbers satisfying the unitary analogue of superior highly composite numbers are a subset of the numbers $n$ such that $\omega(k)<\omega(n)$ for all $k<n$, which are the primorials, products of the first $n$ primes.
Sorry, I was hasty before. I can probably write out a proof tomorrow afternoon, but it is already clear that numbers produced by your recipe are all the primorials and nothing else. I made an unimportant switch to $$ \frac{e^{\omega(n)}}{n^\epsilon} $$
For $\epsilon = 2.0,$ the winner is $n=1.$
For $\epsilon = 1.0,$ the winner is $n=2.$
For $\epsilon = 0.7,$ the winner is $n=6.$
For $\epsilon = 0.6,$ the winner is $n=30.$
For $\epsilon = 0.5,$ the winner is $n=210.$
For $\epsilon = 0.4,$ the winner is $n=2310.$
For $\epsilon = 0.38,$ the winner is $n=30030.$
For $\epsilon = 0.35,$ the winner is $n=510510.$
Note that I actually calculated $\omega(n) - \epsilon \log n.$ You see how it does not matter about $e$ against $2,$ I could calculate $ A \omega(n) - \epsilon \log n$ and simply replace it by $ \omega(n) - (\epsilon / A) \log n$
EDIT: proof easier than usual. It is the primorial with all prime factors satisfying $$ p \leq e^{1 / \epsilon}. $$
At the same time, the first (largest) $\epsilon$ that includes a given prime $p$ is $$ \epsilon = \frac{1}{\log p}. $$ which is irrational.